exam1_solutions

exam1_solutions - Exam
1
for
PHYSICS
2210
...

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Unformatted text preview: Exam
1
for
PHYSICS
2210
 9:30
to
10:45
am
on
Oct
1.
2009
 
 Name:






























































































Score:
 
 
 
 1)
Multiple‐choice
questions
(30
points
total,
and
5
points
each
question).
 
 1A)
For
two
vectors
a
and
b,
which
of
the
following
statement
is
most
accurate?
 
 .
(a
x
b)
=
0.
 A) a
 B) (2a+b)
.
(a
x
b)
=
0.
 C) (a­2b)
.
(a
x
b)
=
0.
 D) A),
B)
and
C)
are
all
true.
 
 
 Answer:
D)
 
 1B)
A
particle
circles
clockwise
around
the
origin
at
a
constant
speed
|vo|.
In
plane
 polar
coordinates,
which
expression
for
r,
v,
and
a
is
correct?
Note
that
r=|r|.
 
 x2 eθ er .P r θ A)
r=rer,
v=­|vo|eθ+|vo|er,
a=|vo|2/r
er.
 B)
r=rer,
v=­|vo|eθ+|vo|er,
a=‐|vo|2/r
er.
 vo C)
r=rer,
v=­|vo|eθ,
a=|vo|2/r
er.
 x1 O D)
r=rer,
v=­|vo|eθ,
a=‐|vo|2/r
er.
 E)
r=rer,
v=|vo|eθ,
a=0.
 
 
 
 Answer:
D)
 
 
 
 
 
 
 
 
 1
 
 1C)
Consider
the
rotational
velocity
field
on
a
plane
v=ωr
eθ.

We
already
showed
 that ∇ × v = 2ωe z 
where
ez
is
unit
vector
perpendicular
to
the
plane.
 L1 = ∫ v ⋅ dr and

 L 2 = ∫ v ⋅ dr are
two
line
integrals
along
circles
Bs1
(solid)
and
Bs2
(dashed),
 
 € BS1 BS 2 € respectively,
where
the
two
circles
have
the
same
areas.

 € 
 
 y 
 |v|=ωr Bs1 o Bs2 A) B) C) D) 
 
 L1>L2.
 L1=L2.
 L1<L2.
 Not
enough
information
to
tell.
 x 
 
 Answer:
B)
 Since
the
curl
is
a
constant
and
two
circles
have
the
same
areas,
using
the
 Stokes’
theorem,
L1
=
L2.
 
 
 1D)
 A
 pendulum
 is
 suspended
 in
 an
 elevator
 that
 is
 descending
 with
 constant
 acceleration
a
( ).
There
are
three
observers
who
are
studying
the
motion
of
the
 pendulum:
observer
A
in
the
elevator
with
the
pendulum,
observer
B
on
the
ground,
 and
observer
C
in
another
elevator
ascending
at
a
constant
velocity.

 
 To
whom
does
the
pendulum’s
motion
satisfy
the
Newton’s
laws?
 
 A) Observer
A.
 B) Observer
B.
 C) Observer
C.
 D) Observers
A,
B,
and
C
(i.e.,
Newton’s
laws
are
always
true).

 E) Observers
B
and
C.
 
 Answer:
E)
 
 
 
 
 2
 1E)
Suppose
that
a,
b
and
c
are
constant,
the
differential
equation:
 a 
 A)
linear,












B)
nonlinear,














C)
homogeneous,










D)
inhomogeneous,




 E)
1st
order,







F)

2nd
order,














G)
Separable
variable.
 € 
 Circle
all
the
properties
that
apply
to
the
equation.
 
 
 Answer:
ADE
 
 
 
 
 1F)
For
a
particle
in
a
conservative
force
field,
which
of
the
following
statement
is
 true?
 
 A)
The
particle’s
linear
momentum
is
always
conserved.
 B)
The
particle’s
angular
momentum
is
always
conserved.
 C)
The
particle’s
total
energy
is
always
conserved.

 D)
The
particle’s
linear
momentum
and
angular
momentum
are
always
conserved.

 E)
The
particle’s
linear
momentum,
angular
momentum
and
total
energy
are
always
 conserved.

 
 Answer:
C)
 
 A)
is
only
true,
if
the
net
force
is
zero,
and
B)
is
only
true,
if
the
net
torque
is
 zero.
 dy + by = ct 2 ,
is:
 dt 
 3
 2)
(20
points)
Consider
a
2‐dimensional
force
field
given
in
a
polar
coordinate
 1 system:
 F = e r .
Is
this
a
conservative
force?
Explain
your
answer.

 r 
 y € o x € 
 
 Answers:

 
 This
force
field
is
a
conservative
force.
 x y e r = cosθe x + sin θe y = e x + e y .
 r r 1 x y x y Therefore,
 F = e r = 2 e x + 2 e y = 2 e+2 e y = Fx e x + Fye y .
 2x r r r x +y x + y2 ∂Fy ∂Fx -2xy 2xy ∇×F =( − )e z = [ 2 +2 ]e z = 0 .

 22 ∂x ∂ y (x + y ) (x + y 2 ) 2 This
proves
for
the
conservative
force.
 € € 
 4
 3)
(20
points).
Consider
an
initially
stationary
particle
with
mass
m
moving
in
a
long
 pipe
 (i.e.,
 1‐dimension
 problem).
 The
 force
 acting
 on
 the
 particle
 to
 move
 the
 particle
forward
decays
with
time
t
as
a
exponential
function
of
mbexp(‐at),
where
a
 and
 b
 are
 constant
 and
 positive.
 The
 particle
 experiences
 a
 retarding
 force
 mkv,
 where
k
is
a
constant
and
v
is
the
block’s
speed.
Write
down
the
Newton’s
2nd
law
in
 a
 convenient
 coordinate
 system
 you
 choose
 (draw
 the
 coordinate
 system),
 and
 determine
how
the
velocity
of
the
particle
varies
with
time.
Assume
that
k>a.
 
 v 
 
 Answers:

 Using
the
above
coordinates,
the
Newton’s
2nd
law
is
 d(ve x ) 
,
 (F − Fr )e x = m dt dv dv mbe− at − mkv = m 

or
 + kv = be− at .
 dt dt st
order
linear
ODE.
 This
is
a
1 Let
P=k,
and
 Q = be− at 
.
 1 € v(t) = ∫ kdt [ ∫ be− at e ∫ kdt dt +C] = e-kt ∫ be (k − a)t dt +Ce-kt ,
 e b -at v(t)€ = e + Ce-kt ,
 k −a where
 C
 is
 a
 constant
 and
 is
 determined
 to
 be
 C=‐b/(k‐a)
 by
 initial
 condition
 v(t=0)=0.
 Therefore,

 b v(t) = (e-at − e-kt ) . k −a o Fr F x € € € € € 
 5
 4)
(30
points).
A
 block
of
mass
 m
with
zero
initial
velocity
slides
down
an
inclined
 plane
for
a
distance
 d
under
the
influence
of
gravity
and
then
coasts
along
the
level
 ground.
 The
 only
 resisting
 force
 on
 the
 block,
 Fr,
 is
 given
 by
 ,
 where
 k
 is
 a
 constant
 and
 v
 is
 the
 block’s
 speed
 (there
 is
 no
 need
 to
 consider
 basal
 friction
 separately).
The
angle
of
inclination
of
the
plane
is
 .
a)
What
is
the
block’s
speed
at
 the
bottom
of
the
inclined
plane?
b)
At
what
time
after
 the
 block
 passes
the
bottom
 of
the
inclined
plane
is
the
 block’s
speed
reduced
to
the
half
of
its
original
speed
at
 the
bottom
of
the
inclined
plane?

 y N o Fr y N x mg o Fr x € € € € € mg 
 Answers:

 a)
Using
the
above
coordinates,
on
the
incline,
the
Newton’s
2nd
law
is
 d(ve x ) dv 
,
or

 (gsinθ − kv 2 ) = .
 (mgsinθ − mkv 2 )e x = m dt dt Because we are given the travel distance d, using the following expression: dv dv dx dv dv leads to (gsinθ − kv 2 ) = v .
 = =v , dt dx dt dx dx € This
equation
is
separable:
 v dx = dv .
 (gsinθ − kv 2 ) € Integrating
this
equation
leads
to

 1 − ln(gsinθ − kv 2 ) = x + C ,
 2k 1 where
C
is
determined
to
be
 C = − ln(gsinθ ) 
using
initial
condition
of
zero
speed.
 2k gsinθ v= (1 - e-2kx ) .

 k gsin€ θ For
x=d,
 v b = (1 - e-2kd ) .
 k dv b)
On
the
horizontal
section,
the
Newton’s
2nd
law
is:
 −kv 2 = ,
whichis
a
 dt vb separable
equation.
We
can
directly
solve
this
equation
for
v(t),
 v = ,
where
 € v bkt + 1 1 € we
considered
v(t=0)=vb.

At
 t = ,
v
is
half
of
vb.
 v bk € 
 € 6
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